Problem Solving in Automata, Languages, and Complexity

(Dana P.) #1

318 COMPUTATIONAL COMPLEXITY


Conversely, let L E AKPA GE(n). W e may assume that there is a one-
tape NTM M1 that accepts L whose accepting computation path does not


read any symbol beyond the two blanks surrounding the input u). (A one-

worktape NTM A& which operates within space n can be easily converted to

an equivalent one-tape NTM A&, using two tracks to simulate the two tapes

of Mz.) We need to construct a context-sensitive grammar to simulate the

computation of Ml.

In Theorems 4.19 and 4.20, we showed how to construct, from a given

DTM M, an unrestricted grammar G such that L(G) = L(M). It is easy to


verify that the same construction works for an NTM M. In the following,

we further convert the grammar corresponding to NTM M, to an equivalent

context-sensitive grammar. First, for each pair (a, a) E Q x C, we define a

nontermial symbol q#a. (For the sake of readability, we write I q#a for the

symbol q#a.) Then, we define a grammar Gr with the following rules:

(A) For each instruction (p, b, L) E S(q, a) of Ml, G1 has the rules


(B) For each instruction (p, b, R) E 6(q, a) of Ml, G1 has the rules

That is, the grammar G1 is just the grammar GM~ of Theorem 4.19 without

the boundary symbols [ and 1, where rules in group (A) correspond to rules

in groups (1) and (2) of GM,, and rules in group (B) correspond to rules in

group (3) of GM~. We note that, since i’& never reads any symbol beyond

the leftmost blank or the first blank to the right of the input, the boundary

symbols are unnecessary in our case. In other words, the second type of rules

in group (3) of GM,, p/J + b pp#B], will never be used.

In addition, in the proof of Theorem 4.19, the second type of rules in

!wUP (a), c pi&q1 + II p#c ] (if (p,B, R) E S(q, a)), is to allow the tape


configurations of the machine &!I to shrink. We eliminate these rules in

grammar G1 here. Thus, Gr still simulates the computation of A& except

that its sentential forms keep all trailing blanks we have ever visited.

From the above analysis, we can see that

(4, XEY) CL, (P, &i) i an f d only if x El q#u y 3 G1 2’ El p#b y’,

where y’ retains the trailing blanks.

Now, we define grammar G2 as follows: It includes all reversals of the rules

of Gr, plus the following rules:

S __) B[h#BIT, T+BTIB,

i-z-l s B ----+Lh, a Lh __) Lh a,
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